Hello guys,

In this simple pi network :

The transfer function is :

Vo/Vi = 1 / (1 + R2/R3 )

R1 is not included in the formula, so Can you explain to me what's the use of R1?
I mean if we removed R1 from the circuit we will get the same result.
Other thing is that if we set R1 to zero then it will become a short circuit and Vo will equal to zero but if we used that formula then setting R1 to zero won't change anything.

So end of the line : What effect does R1 add to the Pi Network ?

Best Regards

 

If you were to omit R1, the voltage out will remain the same as compared to the circuit with it. The reason for this is that R2 & R3 form a voltage divider, so the voltage out is dependent on the ratio of those two resistors.

Now, we know that devices in parallel have the same voltage across them. Applying that to this circuit, we can see that R1 is parallel to the voltage divider(R2 & R3), meaning they have the same voltage impressed across them. This means the voltage that the voltage divider is dividing is the same as that across R1, so R1 doesn't affect the voltage. R1 does, however, affect the current going through that voltage divider...

 

Hello guys,

In this simple pi network :

The transfer function is :

Vo/Vi = 1 / (1 + R2/R3 )

R1 is not included in the formula, so Can you explain to me what's the use of R1?
I mean if we removed R1 from the circuit we will get the same result.
Other thing is that if we set R1 to zero then it will become a short circuit and Vo will equal to zero but if we used that formula then setting R1 to zero won't change anything.

So end of the line : What effect does R1 add to the Pi Network ?

Best Regards

The input source, in general, has finite impedance. Would this make R1 significant?
One use of a pi network is as an impedance matching attenuator. R1 is needed in this case.